Let $$a, b,$$ and $$c$$ denote the sides of a right triangle such that $$c^2 = a^2 + b^2$$. We are interested in triangles where the legs differ by 1, i. e., $$c^2 = a^2 + (a + 1)^2$$. Let's recast the equation so all numbers differ from $$a$$. Thus, $$(a + m)^2 = a^2 + (a + 1)^2$$ Regrouping and solving for $$a$$ we get $$a = (m - 1) \pm \sqrt{2m(m - 1)}$$ Clearly, an integer solution for $$a$$ exists only when $$2m(m - 1)$$ is a perfect square. It's not hard to find the first few values of $$m$$ that yield a perfect square for $$2m(m - 1)$$. They are 1, 2, 9, 50, and 289. Here is a table showing the values for $$a, b$$ and $$c$$ for the given values of $$m$$.

 m a b c 1 0 1 1 2 3 4 5 9 20 21 29 50 119 120 169 289 696 697 985

I noticed two things of interest about this table. The first is that after the first row, each $$m$$ is the sum of $$b$$ and $$c$$ from the previous row. So the next rows in the table would be:

 1,682 4,059 4,060 5,741 9,801 23,660 23,661 33,461 57,122 137,903 137,904 195,025 332,929 803,760 803,761 1,136,689 1,940,450 4,684,659 4,684,660 6,625,109

The second thing I noticed is that each $$m$$ is 1 more than the index of a corresponding square triangle. For example, $$T_{288} = 41,616$$ which is also a perfect square, $$204^2$$. This makes sense because we were looking for values of $$m$$ such that $$2m(m - 1)$$ is a perfect square, and that is the same as looking for values of $$m$$ such that $$4T_{m - 1}$$ is a perfect square. So for every square triangle there is a Pythagorean triple such that $$a$$ and $$b$$ differ by 1, and for every Pythagorean triple where $$a$$ and $$b$$ differ by 1, there is a square triangle.